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AC Drives

Dr. Adel A. El-Samahy

Department ofElectricalEngineering

University of Helwan

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Facts in Electrical Machines If an e.m.f. source is applied to closed

circuit, an electric current will pass through

the circuit When electric current passes through a

conductor or coil, a magnetic field isestablished (DC current produces DC

magnetic field while AC current producesAC magnetic field). The direction of themagnetic field is determined by Right Hand

Grip Rule .

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Facts in Electrical Machines

When a magnetic field cuts a conductor/coil, ane.m.f. is induced in that conductor/coil. Themagnitude of this e.m.f. proportional to the rate ofcutting. The rate of cutting may be produced dueto relative motion between conductors andmagnetic field (DC or AC generators), or AC

magnetic field and stationary winding(transformer) To produce electromagnetic torque, there must be

two magnetic fields. These fields must have no

relative speed and have an angle between them

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Induction Motor Drives

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Importance of Induction motors The induction machine is used in a wide variety of

applications as a means of converting electric powerto mechanical work. It is without doubt theworkhorse of the electric power industry. Pump, steelmill, and hoist drives are but a few applications oflarge multiphase induction motors.

On a smaller scale, the 2-phase servomotor is usedextensively in position-follow-up control systems andsingle-phase induction motors are widely used inhousehold appliances as well as in hand and benchtools

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Advantages of Induction motor

Drives AC motor are less expensive

Ac motors have low maintenance For the same rating, ac motors are higher inweight as compared to dc motors.

AC motors can work in hazardous areas like

chemical, petrochemical etc. whereas dcmotors are unsuitable for such environmentsbecause of commutator sparking

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Disadvantages of Induction

motor Drives

Power converters for1. the control of ac motors are more complex.2. ac motors are more expensive3. ac motors generate harmonics in the supply

system and load circuit. Hence AC motors getsderated.

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Induction Motor Drives?Why does the rotor rotates? When the stator winding is connected to 3 phase AC

supply, a resultant rotating magnetic field withconstant magnitude is established. This field rotates

with synchronous speed = 12

Where:fis the supply frequency,Pis the number of poles

This field cuts the rotor winding which is short

circuited inducing an e.m.f. that causes current in therotor winding which in turns produces another rotatingmagnetic fields.

The interaction between the two fields produces an

electromagnetic torque that forces the rotor to rotate

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Induction Motor Drives?Definitions

ns: Speed of rotating magnetic field (Synchronous speed) rpmws: Speed of rotating magnetic field (Synchronous speed) rad/s

n: motor speed rpmw: rotor speed rad/sf1: supply frequencyf2: frequency of the induced e.m.f in the rotor = sf1s : slipns- n: Slip speed

s

s

s

s

n

nns

www

p

fns

112060

2 ss

nw

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Induction Motor Drives?

Example 1: . Calculate the frequency of the rotor current to

produce an average torque at a speed of - 200 rpm.(p=2 and f1=50 Hz)

Solution

rpmp

fns 3000

120 1

1.06673000

)200(3000

s

s

n

nns

f2 = sf1= 53.33 Hz

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Induction Motor Equivalent Circuit

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`212

`

21

`

2

XX

s

RR

VI

ph

tan-1((X1+X2

)/R1+R2/s))

I1=Io+ I2Im+ I2

Where: Io is no load current

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Power Flow Graph

Input Power = Pin= 3VphIphcos f=3VlIlcos fStator copper losses = Pst= 3Iph

2R1Core losses = Pc= 3Vm

2/RmAir gap power = P

g

= 3I2

2R2

/sMechanical Power (developed power) = Pm=Pg(1-s) = 3I2

2R2/s(1-s)Pf=friction lossesOutput Power = P0= PmPf= Po/Pin

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Power Flow Graph

Example 2:. Find the efficiency of 3-phase, 50 Hp, 60 Hz,

four-pole induction motor. The motor is star connected.nf.l=1755rpm, VL= 440 V, Io= 18 A. at 0.085 Power factor lag.

Also available from the manufacturers data sheet are the

following design data: R1=0.1,

R2'=0.12 , X1=0.35 , X2

'=0.4 , Pcore=1200 W, Prot=950 W.

Solution

=1800 rpm

pfns1120

025.01800

)1755(1800

s

s

n

nns

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Power Flow GraphExample 2:

Pin= 3 440 57.35 cos (26.7) = 39046 W

Pg= 3 I2'2 r2

'/s =37896 W

Pout= Pm- Prot= Pg(1-s)-950 =36000 W

= Pout/

Pin

= 92.2 %

Io= Im+ Irot =18 -850 Irot =950/(3 440)

Im= 0.35 +j 17.95

I2'

=Vph/Z2

I1= Im+ I2' = 57.35 -26.70

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Torque Speed Characteristics

sXXs

RR

RVT

s

RIPT

s

ph

ss

g

2`

21

2`

2

1

`

2

2

`

2

2`

2

`

3

3

w

ww

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Regenerativebraking

Pluging

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Torque Speed CharacteristicsThere are three region of operation:1. Motoring (0s1)

The motor rotates in the same direction asthe field; as the slip increases, the torqueincreases, while air gap flux remains constant.Once the torque reaches its maximum value

(s = scr), the torque decreases with the increase inthe slip due to reduction in the air gap flux

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There are three region of operation:2.Regeneration (s0)

The speed is greater than the synchronousspeed, wand wsbeing in the same direction andthe slip is negative. Therefore R2/s is negativewhich means the power is fed back from the shaft

to the supply. The torque speed characteristics issimilar to that of motoring, but having negativevalue

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There are three region of operation:3.Plugging (1 s)

The speed is opposite to the direction of thefield, this may happen if the sequence of thesupply source is reversed, while motoring,therefore the direction of the field and the

developed torque also reversed. The energy due toplugging must be dissipated within the motor andmay cause excessive heating.

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Torque Speed CharacteristicsCondition for Maximum Torque

dT/ds = 0

2`

21

2

11

2

2`

21

2

1

`2

2

3

XXRR

VT

XXR

Rs

s

phmm

cr

w

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2`21

211

2

2

3

XXRR

VT

s

ph

mr

w

The maximum regenerative torque

The maximum regenerative torque may be obtained from the

general torque equation by letting s= -scr

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2211

2211

2

`2

1

211

12

2

12

cr

cr

mmmr

cr

crcrmmst

crcr

cr

crmm

XRR

XRRTT

as

ssaTT

RRa

sas

s

s

s

asTT

`

21 XXXcr

s

s

s

s

TT

cr

cr

max

2

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s

s

s

s

TT

cr

cr

mm

2

R1 is small compared to the other circuit impedances formotor with rating greater than 1 kW and my beneglected.

In this case

`21

2

2

3

XX

VTT

s

ph

mrmm

w

`

21

`

2

XX

Rscr

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